A rock is dropped off the edge of a cliff and its distance s (in feet) from the

Question

A rock is dropped off the edge of a cliff and its distance s (in feet) from the top of the cliff after t seconds is s(t) = 16t^2. Assume the distance from the top of the cliff to the water below is 1936 ft. Answer part (a) and (b). (a) When will the rock strike the water? The rock will strike the water at 11 seconds. (Type an integer or a simplified fraction.) (b) Make a table of average velocities and approximate the velocity at which the rock strikes the water. (Type an integer or decimal rounded to three decimal places as needed) The rock strikes the water at approximately ft/sec. (Round to the nearest integer as needed)

Explanation / Answer

16t² = 1936

t² =121

t=121

t= 11 second

Now,

s =16t²

v = 32t

v= 32(11)

=352

which is also obvious from the last column of table.

Answer will be

352 ft/sec

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